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3 years ago

What is the range of the quadratic function?

Mathematics

1 answer:

tigry1 [53]3 years ago

4 0

For every polynomial function (such as quadratic functions for example), the domain is all real numbers. if the parabola is opening upwards, i.e. a > 0 , the range is y ≥ k ; if the parabola is opening downwards, i.e. a < 0 , the range is y ≤ k .

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How tooo do thissss?

exis [7]

Step-by-step explanation:

m/s² = m/s / s

this is the slope of the line in the graph.

t = 2, that means it is in the first segment of 0 - 6 seconds

the slope in this segment is

(25-15)m/s / (6-0)s

10m/s / 6s = 10/6 m/s² = 5/4 m/s² = 1.25 m/s² ≈ 1.3 m/s²

the same thing for t = 12 in the third segment.

the acceleration or slope is negative here, but that's called here a "forbidden word", and is therefore also a positive number.

in that this segment the slope is

(0-25)m/s / (16-11)s

-25m/s / 5s = -5 m/s²

and so, the answer is 5 m/s².

for the total distance we need to calculate the area under the lines.

the line functions in the 3 segments are :

s(t) = 5/4 t + 15

s(t) = 25

s(t) = -5t + 25

the third one is simplified by moving it from the 11 to 16 seconds interval to a 0 to 5 seconds interval. this creates the same area under that line.

the area under the first line is the integral of the line function between 0 and 6.

that is

[5/8 t² + 15t] between 0 and 6.

and that is

5/8 × 36 + 90 - 5/8 × 0 - 15×0 = 45/4 + 90

= 45/4 + 360/4 = 405/4 = 101.25m

the area under the second segment is simply the area of the rectangle 5×25 = 125m

and the area under the third segment is the integral of the line function between 0 and 5 (remember, we shifted from 11 to 16 to the interval 0 to 5).

that is

[-2.5t² + 25t] in the interval 0 to 5.

and that is

-2.5×25 + 25×5 - -2.5×0 - 25×0 = -62.5 + 125 = 62.5m

as a control check this area is also the area of a right-angled triangle with legs of 5 and 25

the area is 25×5/2 = 125/2 = 62.5

so, it is correct.

the total area under the lines is therefore

101.25 + 125 + 62.5 = 288.75 m ≈ 288.8 m

the average speed is now simply

288.75 m / 16 s = 288.75/16 m/s = 18.046875 m/s ≈

≈ 18 m/s

but if we used the already rounded result of the total distance of 288.8 m, then we would get

288.8 / 16 = 18.05 m/s ≈ 18.1 m/s

3 0

3 years ago

A group of n students is assigned seats for each of two classes in the same classroom. how many ways can these seats be assigned

Ratling [72]

Consider ONLY the first class. Since there are n seats (assuming they are in a row), we would have n! ways.

Now, consider the second class. Let's start by arranging three people first and then generalising n people to better understand what is going on.

Where we have 3 people:
First class: A B C = 3!
Second class _ _ _
Now, consider where each of these people can't actually sit.
For A, it's the first seat, for B, it's the second seat, and for C, it's the last seat.
This means that we have a restriction on EACH of the candidates.

So, to tackle this, let's consider A only; B and C will follow the same way.
A can sit in 2 different spots, namely the second and last seat: thereby, having 3C2 ways in sitting. _ A _
Now, when we fix one person, C can only go in one place: first seat. This means that for one single arrangement of the first class, we've made: 3C2 arrangements for the second class, for ONE particular person. Extend that to another person, and we get: 2C1 ways

This extends to what we call: a derangement where the number of permutations made contains no fixed element. We can regard things like picking up three/two/one pen as derangements, because really we're arranging AND not arranging them simultaneously.

Thus, we use the inclusion-exclusion method:
Total no. of perms:
$n! \cdot n!\left(1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - ... + (-1)^{n} \cdot \frac{1}{n!}\right)$